a) Ta có: \(B{C^2} = {\overrightarrow {BC} ^2} = {\left( {\overrightarrow {AC} - \overrightarrow {AB} } \right)^2}\)\( = {\overrightarrow {AC} ^2} + {\overrightarrow {AB} ^2} - 2\overrightarrow {AC} .\overrightarrow {AB} \)
Do đó \(\overrightarrow {AB} .\overrightarrow {AC} = \dfrac{{{{\overrightarrow {AC} }^2} + {{\overrightarrow {AB} }^2} - {{\overrightarrow {BC} }^2}}}{2}\)\( = \dfrac{1}{2}\left( {{8^2} + {6^2} - {{11}^2}} \right) = - \dfrac{{21}}{2}\)
\( \Rightarrow AB.AC.cosA = - \dfrac{{21}}{2}\) \( \Rightarrow \cos A = - \dfrac{7}{{32}} < 0\) nên \(A\) là góc tù.
b) Ta có \(\overrightarrow {AM} = \dfrac{1}{3}\overrightarrow {AB} ;\overrightarrow {AN} = \dfrac{1}{2}\overrightarrow {AC} \)
Do đó \(\overrightarrow {AM.} \overrightarrow {AN} = \dfrac{1}{3}\overrightarrow {AB} .\dfrac{1}{2}\overrightarrow {AC} \)\( = \dfrac{1}{6}\overrightarrow {AB} .\overrightarrow {AC} = \dfrac{1}{6}.\left( { - \dfrac{{21}}{2}} \right) = - \dfrac{7}{4}\)
